# Metric Spaces and Compactness

1. Sep 28, 2010

### Buri

My professor in lecture yesterday said that if a set is closed and bounded in a metric space it doesn't necessarily imply that it is compact. If X = R^n, then it does happen to be true, however. I was trying to construct an example, but I am getting confused. If I let X = R, and Y = (0,1) where Y is a subspace of X, then A = (0,1/2] is closed and bounded in Y. However, from where do I choose the open cover? That is, open relative to X or Y? I know in this case it won't make a difference, but maybe in differently chosen X, Y and A it might. I guess this is a matter of definition, but would like some help.

Thanks a lot.

2. Sep 28, 2010

### Office_Shredder

Staff Emeritus
Don't think about a subset of Rn. Metric spaces just are, they aren't normally viewed as being in a larger metric space even if that is useful. In this case you have X=(0,1) with the Euclidean metric, and A=(0,1/2]. The fact that you can embed this metric space into the real numbers is irrelevant

3. Sep 28, 2010

### Buri

Ahh I see. Thanks!